1

CHEMISTRY

1. A stream of electrons from a heat filament was passed between two charge plates kept at a potential

difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h

λ (where λ is

wavelength associated with electron wave) is given by:

(1) 2 me V

(2) √meV

(3) √2 meV

(4) me V

Solution: (3)

Given stream of electron from heated filament was passed between two charge plates at potential difference V

e, m are charge and mass of electron

V =E

e

eV =1

2× m × V2

λ =h

mv V =

h

mλ

eV =1

2× m × [

h

mλ]

2

h

λ= √2 meV

2. 2 – chloro – 2 – methylpentane on reaction with sodium methoxide in methanol yields:

(i)

(ii)

(iii)

2

(1) i and iii

(2) iii only

(3) i and ii

(4) All of these

Solution: (4)

3. Which of the following compounds is metallic and ferromagnetic?

(1) CrO2

(2) VO2

(3) MnO2

(4) TiO2

Solution: (1)

d – block elements are metals.

MnO2 exhibit strong attraction to magnetic fields and are able to retain their magnetic properties.

So, it exhibits metallic character and it’s ferromagnetic.

4. Which of the following statements about low density polythene is FALSE?

(1) It is a poor conductor of electricity

(2) Its synthesis required dioxygen or a peroxide initiator as a catalyst

3

(3) It is used in the manufacture of buckets, dust – bins etc.

(4) Its synthesis requires high pressure

Solution: (3)

Low density polythene: It is obtained by the polymerization of ethane high pressure of 1000-2000 atm. at a temp.

of 350 K to 570 K in the pressure of traces of dioxygen or a peroxide initiator (cont).

Low density polythene is chemically inert and poor conductor of electricity. It is used for manufacture squeeze

bottles. Toys and flexible pipes.

5. For a linear plot of log (x

m) versus log p in a Freundlich adsorption isotherm, which of the following

statements is correct? (k and n are constants)

(1) 1

n appears as the intercept

(2) Only 1

n appears as the slope

(3) log (1

n) appears as the intercept

(4) Both k and1

n appear in the slope term

Solution: (2)

According to Freundlich isotherm

x

m= k. p

1

n

logx

m= log k +

1

nlog P

So intercept is log k and slope is 1

n

6. The heats of combustion of carbon and carbon monoxide are −393.5 and − 283.5 kJ mol−1, respectively.

The heat of formation (in kJ) of carbon monoxide per mole is:

(1) 676.5

(2) − 676.5

(3) −110.5

(4) 110.5

Solution: (3)

Given heat of combustion of carbon and carbon monoxide are −393.5 and − 283.5 kJ mol−1, respectively

(i) C + O2 → CO2 ∆H1 = −393.5 kJ mol−1

(ii) CO +1

2O2 → CO2 ∆H2 = −283.5 kJ mol−1

4

To find heat of formation of CO per mole.

i.e. C +1

2O2 → CO

(ii) × (2)

2CO + O2 → 2CO2 ∆H3 = −283.5 × 2 kJ mol−1

2CO2 → 2CO + O2 ∆H4 = 567 kJ mol−1

(i) × (2)

2C + 2O2 → 2CO2 ∆H5 = −393.5 × 2 kJ mol−1

= −787 kJ mol−1

2C + O2 → 2CO ∆H6 = −220 kJ mol−1

For one mole of CO,

∆H =−220

2= −110 kJ mol−1

7. The hottest region of Bunsen flame shown in the figure below is:

(1) Region 2

(2) Region 3

(3) Region 4

(4) Region 1

Solution: (1)

5

8. Which of the following is an anionic detergent?

(1) Sodium lauryl sulphate

(2) Cetyltrimethyl ammonium bromide

(3) Glyceryl oleate

(4) Sodium stearate

Solution: (1)

(i) Cetyltrimethyl ammonium bromide

(ii) Glyceryl oleate

(iii) Sodium stearate

(iv) Anionic surfactant

9. 18 g glucose (C6H12O6) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous

solution is:

6

(1) 76.0

(2) 752.4

(3) 759.0

(4) 7.6

Solution: (2)

∆P

Po = mol. Fraction of glucose

760−PSoln

760=

W1Mwt1

W1M.wt1

+w2

M.wt2

=18

18018

180+

178.2

18

=0.1

0.1+9.9=

1

100

760 − PSoln =760

100

PSol = 752.4

10. The distillation technique most suited for separating glycerol from spent – lye in the soap industry is:

(1) Fractional distillation

(2) Steam distillation

(3) Distillation under reduced pressure

(4) Simple distillation

Solution: (3)

Glycerol (B.P. 290oC) is separated from spent – lye in the soap industry by distillation under reduced pressure, as

for simple distillation very high temperature is required which might decompose the component.

11. The species in which the N atom is a state of sp hybridization is:

(1) NO2−

(2) NO3−

(3) NO2

(4) NO2+

Solution: (4)

N SP

(i)

7

(ii)

(iii)

(iv)

12. Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases

from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of

formation of O2 will be:

(1) 6.93 × 10−4 mol min−1

(2) 2.66 L min−1 at STP

(3) 1.34 × 10−2 mol min−1

(4) 6.93 × 10−2 mol min−1

Solution: (1)

2H2O2 → 2H2O + O2

[O2] =[H2O2]

2=

[H2O]

2

For, t1

2

, H2O2 decreases to 0.125M from 0.5M

So, 2 × t1

2

= 50

t1

2

= 25

t1

2

=0.69314

K

K =0.69314

25

[O2] =1

2×

0.69314

25

8

= 6.93 × 10−4 mol min−1

13. The pair having the same magnetic moment is:

[At. No. ∶ Cr = 24, Mn = 25, Fe = 26, Co = 27]

(1) [Cr(H2O)6]2+ and [Fe(H2O)6]2+

(2) [Mn (H2O)6]2+ and [Cr(H2O)6]2+

(3) [CoCl4]2− and [Fe(H2O)6]2+

(4) [Cr(H2O)6]2+ and [CoCl4]2−

Solution: (1)

In option A: [Mn(H2O)6]2+(3d5) with

& [Cr(H2O)6]2+, Cr2+(3d4) with W.F.L.,

In option B: [CoCl4]2−, Co2+(3d7) with W.F.L.,

& [Fe(H2O)6]2+, Fe2+(3d6) with W.F.L.,

In option C: [Cr(H2O)6]2+, Cr2 + (3d4) with W.F.L.,

9

& [CoCl4]2−, Co2+ (3d7) with W.F.L.,

In option D: [Cr(H2O)6]2+, Cr2+(3d4) with W.F.L.,

& [Fe(H2O)6]2+, Fe2+(3d6) with W.F.L.,

Here both complexes have same unpaired electrons i.e. = 4

14. The absolute configuration of

is:

(1) (2S, 3R)

(2) (2S, 3S)

(3) (2R, 3R)

(4) (2R, 3S)

Solution: (1)

10

∴ Configuration is 2S, 3R.

15. The equilibrium constant at 298 K for a reaction A + B ⇌ C + D is 100, If the initial concentration of all the

four species were 1 M each, then equilibrium concentration of D (in mol L−1) will be:

(1) 0.818

(2) 1.818

(3) 1.182

(4) 0.182

Solution: (2)

K at 298 K for the reaction

A + B ⇌ C + D is 100

Given initial concentration of all four species is 1M

At 1 = 0,

∴ A1

+B1

⇌C1

+D1

At equilibrium,

A + B1 − x 1 − x

⇌ C + D

1 + x 1 + x

K =(1+x) (1+x)

(1−x) (1−x)= 100

(1−x)

1−x= 10

x =9

11= 0.818

[D] = 1 + x = 1 + 0.818 = 1.818

11

16. Which one of the following ores is best concentrated by froth floatation method?

(1) Siderite

(2) Galena

(3) Malachite

(4) Magnetite

Solution: (2)

Froth floatation method is used for concentration of sulphide ores.

↓ Galena → Pbs

17. At 300 K and 1 atm, 15mL of a gaseous hydrocarbon requires 375 mL air containing 20% 𝑂2 by volume

for complete combustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in liquid

form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is:

(1) 𝐶3𝐻8

(2) 𝐶4𝐻8

(1) 𝐶4𝐻10

(2) 𝐶3𝐻6

Solution: (Bonus or 1)

Volume of 𝑁2 in air = 375 × 0.8 = 300 𝑚𝑙

Volume of 𝑂2 in air = 375 × 0.2 = 75 𝑚𝑙

𝐶𝑥𝐻𝑦 + (𝑥 +𝑦

4) 𝑂2

15𝑚𝑙 15 (𝑥 +𝑦

4)

0 0

→ 𝑥𝐶𝑂2(𝑔) + 15𝑥

𝑦

2𝐻2𝑂(ℓ)

−

After combustion total volume

330 = 𝑉𝑁2+ 𝑉𝐶𝑂2

330 = 300 + 15𝑥

𝑥 = 2

Volume of 𝑂2 used

15 (𝑥 +𝑦

4) = 5

𝑥 +𝑦

4= 5

𝑦 = 12

12

So hydrocarbon is = 𝐶2𝐻12

None of the option matches it therefore it is a BONUS.

Alternatively

𝐶𝑥𝐻𝑦 + (𝑥 +𝑦

4) 𝑂2

15𝑚𝑙 15 (𝑥 +𝑦

4)

0 0

→ 𝑥𝐶𝑂2(𝑔) + 15𝑥

𝑦

2𝐻2𝑂(ℓ)

−

Volume of 𝑂2 used

15 (𝑥 +𝑦

4) = 75

𝑥 +𝑦

4= 5

If further information (i.e., 330 ml) is neglected, option (𝐶3𝐻8) only satisfy the above equation.

18. The pair in which phosphorous atoms have a formal oxidation state of + 3 is:

(1) Pyrophosphorous and hypophosphoric acids

(2) Orthophosphorous and hypophospheric acids

(3) Pyrophosphorous and pyrophosphoric acids

(4) Orthophosphorous and pyrophosphorous acids

Solution: (4)

Acid Formula Formal oxidation state of phosphorous

Pyrophosphorous acid

H4P2O5 +3

Pyrophosphoric acid H4P2O7 +5

Orthophosphorous acid

H3PO3 +3

Hypophosphoric acid H4P2O6 +4

Both pyrophosphorous and orthophosphorous acid have +3 formal oxidation state

19. Which one of the following complexes shows optical isomerism?

(1) cis [Co(en)2 Cl2]Cl

(2) trans [Co(en)2 Cl2] Cl

(3) [Co(NH3)4 Cl2] Cl

(4) [Co (NH3)3 Cl3]

13

Solution: (1)

Geometrical isomers

cis [Co(en)2Cl2]Cl

trans [Co(en)2Cl2]Cl

Cis trans

20. The reaction of zinc with dilute and concentrated nitric acid, respectively, produces:

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(1) NO2 and NO

(2) NO and N2O

(3) NO2 and N2O

(4) N2O and NO2

Solution: (4)

Zn on reaction with HNO3

4Zn + 10HNO3 → 4Zn(NO3)2 + N2O + 5H2O

Zn + 4HNO3(con) → Zn(NO3)2 + 2NO2 + 2H2O

21. Which one of the following statements about water is FALSE?

(1) Water can act both as an acid and as a base

(2) There is extensive intramolecular hydrogen boding in the condensed phase

(3) Ice formed by heavy water sinks in normal water

(4) Water is oxidized to oxygen during photosynthesis

Solution: (2)

(i) Water can show amphiprotic nature and hence water can act both as an acid a base.

(ii) There is extensive intermolecular hydrogen bonding in the condensed phase instead of intramolecular H –

bonding.

(iii) Ice formed by heavy water sinks in normal water due to higher density of D2O than normal water.

(iv)

22. The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was

found to be 1000 ppb, 40 ppb, 100 ppm and 0.2 ppm, respectively. This water is unsuitable for drinking due to

high concentration of:

(1) Lead

(2) Nitrate

(3) Iron

(4) Fluoride

Solution: (2)

Concentration of fluoride = 1000 PPb

= 1 PPm

15

Concentration of lead = 40 PPb

= 0.04 PPm

Concentration of nitrate = 100 PPm

Concentration of iron = 0.2 PPm

High concentration of nitrate

23. The main oxides formed on combustion of Li, Na and K in excess of air are, respectively:

(1) LiO2, Na2O2 and K2O

(2) Li2O2, Na2O2 and KO2

(3) Li2O, Na2O2 and KO2

(4) Li2O, Na2O and KO2

Solution: (3)

In 1A group

Li on r × n with excess air

4Li + O2 → 2Li2O

Na on r × n with excess air

2Na + O2 → Na2O2

K on r × n with excess air

K + O2 → KO2

24. Thiol group is present in:

(1) Cystine

(2) Cysteine

(3) Methionine

(4) Cytosine

Solution: (2)

Thiol group (−SH)

Cysteine

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25. Galvanization is applying a coating of:

(1) Cr

(2) Cu

(3) Zn

(4) Pb

Solution: (3)

Galvanization is the process of applying zinc coating to steel (or) iron, to prevent rusting

26. Which of the following atoms has the highest first ionization energy?

(1) Na

(2) K

(3) Sc

(4) Rb

Solution: (3)

Na is the smallest element in the IA group elements and it has highest IE among K, Rb

Sc has lowest effective nuclear charge

Effective nuclear charge ×1

I.E

Sc has low effective nuclear charge than Na.

So it has I.E. among given elements.

27. In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of

amine produced are:

(1) Four moles of NaOH and two moles of Br2

(2) Two moles of NaOH and two moles of Br2

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(3) Four moles of NaOH and one mole of Br2

(4) One mole of NaOH and one mole of Br2

Solution: (3)

To find number of moles of NaOH and Br2 used per mole of amine produced.

28. Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure pi and temperature T1

are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one

of the bulbs is then raised to T2. The final pressure Pf is:

(1) 2pi (T1

T1+T2)

(2) 2pi (T2

T1+T2)

(3) 2pi (T1T2

T1+T2)

(4) pi (T1T2

T1+T2 )

Solution: (2)

Given two closed bulbs of equal volume (v) containing ideal gas initially of pressure pi and temperature T1 which

are connected by narrow tube of negligible volume.

To find final pressure Pf when one raised to T2

No. of moles of gas doesn’t change

(𝑛𝑇)𝑖= (𝑛𝑇)𝑓

𝑃𝑖𝑉

𝑅𝑇1+

𝑃𝑖𝑉

𝑅𝑇1=

𝑃𝑓𝑉

𝑅𝑇1+

𝑃𝑓𝑉

𝑅𝑇2

2𝑃1

𝑇1=

𝑃𝑓

𝑇1+

𝑃𝑓

𝑇2

18

29. The reaction of propene with HOCl (cl2 + H2O) proceeds through the intermediate:

(1) CH3 − CH+ − CH2 − Cl

(2) CH3 − CH(OH) − CH2+

(3) CH3 − CHCl − CH2+

(4) CH3 − CH+ − CH2 − OH

Solution: (1)

30. The product of the reaction give below is:

(1)

(2)

(3)

19

(4)

Solution: (1)

MATHEMATICS

31. Two sides of a rhombus are along the lines, 𝑥 − 𝑦 + 1 = 0 and 7𝑥 − 𝑦 − 5 = 0. If its diagonals intersect at

(−1, −2), then which one of the following is a vertex of this rhombus ?

(1) (−3, −8)

(2) (1

3, −

8

3)

(3) (−10

3, −

7

3)

(4) (−3, −9)

Solution: (2)

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𝑑1 ∶ 𝑦 − 2 =−2 − 2

−1 − 1 (𝑥 − 1)

𝑦 − 2 = 2(𝑥 − 1)

𝑦 − 2𝑥 = 0

𝑑2 ⊥ 𝑑1 ⇒ 2𝑦 + 𝑥 = 𝑘 𝑝 on (−1, −2)

2𝑦 + 𝑥 + 5 = 0

Non P.O.I. of 𝑑2 and 𝐿1

𝑥 − 𝑦 + 1 = 0

𝑥 + 2𝑦 + 5 = 0

−3𝑦 − 4 = 0

𝑦 = −4

3

And P.O.I. of 𝑑2 and 𝐿2

𝑥 + 2𝑦 + 5 = 0

14𝑥 − 2𝑦 − 10 = 0

And 𝑦 = −8

3

15𝑥 − 5 = 0

⇒ 𝑥 =1

3

32. If the 2𝑛𝑑, 5𝑡ℎ and 9𝑡ℎ term of a non – constant A.P. are in G.P., then the common ratio of this G.P. is:

(1) 4

3

(2) 1

(3) 7

4

21

(4) 8

5

Solution: (1)

Let the A.P. be a, a + d, a + 2d,…..

Given (𝑎 + 𝑑). (𝑎 + 8𝑑) = (𝑎 + 4𝑑)2

𝑎2 + 9𝑎𝑑 + 8𝑑2 = 𝑎2 + 8𝑎𝑑 + 16𝑑2

8𝑑2 − 𝑎𝑑 = 0

𝑑[8𝑑 − 𝑎] = 0

∵ 𝑑 ≠ 0

𝑑 =𝑎

8

So, 2𝑛𝑑 term 5𝑡ℎ term 9𝑡ℎ term

Would be (𝑎 +𝑎

8) (𝑎 +

𝑎

2) (𝑎 + 𝑎)

9𝑎

8

3𝑎

2 2𝑎

Common Ratio : 2𝑛𝑑 𝑡𝑒𝑟𝑚

1𝑠𝑡 𝑡𝑒𝑟𝑚=

3𝑎

2.9𝑎 . 8 =

4

3

33. Let P be the point on the parabola, 𝑦2 = 8𝑥 which is at a minimum distance from the centre C of the circle,

𝑥2 + (𝑦 + 6)2 = 1. Then the equation of the circle, passing through C and having its centre at P is:

(1) 𝑥2 + 𝑦2 − 𝑥 + 4𝑦 − 12 = 0

(2) 𝑥2 + 𝑦2 −𝑥

4+ 2𝑦 − 24 = 0

(3) 𝑥2 + 𝑦2 − 4𝑥 + 9𝑦 + 18 = 0

(4) 𝑥2 + 𝑦2 − 4𝑥 + 8𝑦 + 12 = 0

Solution: (4)

𝑦2 = 8𝑥 is the equation of the given parabola. If P is a point at minimum distance from (0, -6) then it should be

normal to the parabola at P.

Slope of tangent

2𝑑𝑦

𝑑𝑥 . 𝑦 = 8

𝑑𝑦

𝑑𝑥=

4

𝑦.

∴ Slope of normal = (−𝑦

4)

22

Any point on the parabola would be (𝑦2

8, 𝑦), and hence slope of the normal would be

(𝑦 + 6)

𝑦2

8

= −𝑦

4

(𝑦 + 6) = −𝑦3

32

At 𝑦 = −4, LHS = RHS

(2) = +64

32= 2

At 𝑦 = −4 𝑥 = 2

So point 𝑃(2, −4).

Radius of the desired circle would be √22 + 22 = 2√2,

So equation of the circle would be √(𝑥 − 2)2 + (𝑦 + 4)2 = 2√2

𝑥2 − 4𝑥 + 4 + 𝑦2 + 8𝑦 + 16 = 8

𝑥2 + 𝑦2 − 4𝑥 + 8𝑦 + 12 = 0

34. The system of linear equations

𝑥 + 𝜆𝑦 − 𝑧 = 0

𝜆𝑥 − 𝑦 − 𝑧 = 0

𝑥 + 𝑦 − 𝜆𝑧 = 0

Has a non –trivial solution for :

(1) Exactly one value of 𝜆

(2) Exactly two values of 𝜆

(3) Exactly three values of 𝜆

(4) Infinitely many values of 𝜆

Solution: (3)

For non – trivial solution.

Δ = |1 𝜆 −1𝜆 −1 −11 1 −𝜆

| = 0

23

1(𝜆 + 1) − 𝜆(−𝜆2 + 1) − 1(𝜆 + 1) = 0

(𝜆 + 1) − 𝜆(1 − 𝜆) (1 + 𝜆) − (𝜆 + 1) = 0

𝜆(1 − 𝜆) (1 + 𝜆) = 0

𝜆 = 0, 𝜆 = 1, 𝜆 = −1

Exactly there value of 𝜆.

35. If 𝑓(𝑥) + 2𝑓 (1

𝑥) = 3𝑥, 𝑥 ≠ 0, and 𝑆 = {𝑥 ∈ 𝑅 ∶ 𝑓(𝑥) = 𝑓(−𝑥)}; then S :

(1) Contains exactly one element

(2) Contains exactly two elements

(3) Contains more than two elements

(4) Is an empty set

Solution: (2)

𝑓(𝑥) + 2 . 𝑓 (1

𝑥) = 3𝑥 …….(i)

Replace 𝑥 by 1

𝑥

𝑓 (1

𝑥) + 2𝑓(𝑥) =

3

𝑥 ……..(ii)

________________________________

3𝑓(𝑥) =6

𝑥− 3𝑥

𝑓(𝑥) =2

𝑥− 𝑥

∵ 𝑓(𝑥) = 𝑓(−𝑥)

Therefore 2

𝑥− 𝑥 = −

2

𝑥+ 𝑥

4

𝑥= 2𝑥

2 = 𝑥2

𝑥 = ± √2

Contains exactly two elements

36. Let 𝑝 = lim𝑥→0+

(1 + tan2 √𝑥 )1

2𝑥, then log p is equal to :

24

(1) 1

(2) 1

2

(3) 1

4

(4) 2

Solution: (2)

Let p = ln(1 + tan2 √𝑥)1

2𝑥.

The limit is of the form (1 + 0)∝ = 𝑒0 .∝

𝑒tan √𝑥 .tan √𝑥

√𝑥 √𝑥 .12

lim𝑥→0+

𝑝 = 𝑒12

log 𝑝 = log 𝑒12 =

1

2

=1

2

37. A value of 𝜃 for which 2+3𝑖 sin 𝜃

1−2𝑖 sin 𝜃 is purely imaginary, is :

(1) 𝜋

6

(2) sin−1 (√3

4)

(3) sin−1 (1

√3)

(4) 𝜋

3

Solution: (3)

Let 𝑧 =2+3𝑖 sin 𝜃

1−2𝑖 sin 𝜃

Rationalizing the complex number.

(2 + 3𝑖 sin 𝜃) (1 + 2𝑖 sin 𝜃)

1 + 4 sin2 𝜃

=(2 − 6 sin2 𝜃) + 𝑖 (7 sin 𝜃)

1 + 4 sin2 𝜃

To make it purely imaginary. It real part should be ‘O’.

25

Hence 2 = 6 sin2 𝜃

sin 𝜃 =1

√3

sin−1 (1

√3)

38. The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its

conjugate axis is equal to half of the distance between its foci, is :

(1) 4

√3

(2) 2

√3

(3) √3

(4) 4

3

Solution: (2)

2𝑏2

𝑎= 8 …(i)

2𝑏 =1

2 2𝑎𝑒 ….(ii)

𝑏

𝑎=

𝑒

2

From (ii)

Now, 𝑒 = √1 +𝑏2

𝑎2 = √1 +𝑒2

4

𝑒2 = 1 +𝑒2

4

3

4𝑒2 = 1

𝑒2 =4

3= 1

𝑒 =2

√3

39. If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true ?

(1) 3𝑎2 − 32𝑎 + 84 = 0

(2) 3𝑎2 − 34𝑎 + 91 = 0

26

(3) 3𝑎2 − 23𝑎 + 44 = 0

(4) 3𝑎2 − 26𝑎 + 55 = 0

Solution: (1)

𝑥 𝑥2

2 4

3 9

𝑎 𝑎2

11 121

16 + 𝑎 134 + 𝑎2

√∑ 𝑥2

4− (

∑ 𝑥𝑖

4)

2

√134 + 𝑎2

4− (

16 + 𝑎

4)

2

=35

10

1

2√134 + 𝑎2 −

(16 + 𝑎)2

4=

7

2

√536 + 4𝑎2 − 256 − 𝑎2 − 32𝑎 = 7

40. The integral ∫2𝑥12+5𝑥9

(𝑥5+𝑥3+1)3 𝑑𝑥 is equal to:

where C is an arbitrary constant.

(1) 𝑥10

2(𝑥5+𝑥3+1)2 + 𝐶

(2) 𝑥5

2(𝑥5+𝑥3+1)2 + 𝐶

(3) −𝑥10

2(𝑥5+𝑥3+1)2 + 𝐶

(4) −𝑥5

(𝑥5+𝑥3+1)2 + 𝐶

Solution: (1)

∫2𝑥12 + 5𝑥9

(𝑥5 + 𝑥3 + 1)3𝑑𝑥

∫(2𝑥12 + 5 + 9) 𝑑𝑥

𝑥15 (1 +1

𝑥2 +1

𝑥5)3

27

∫(

2𝑥3 +

5𝑥6) 𝑑𝑥

(1 +1

𝑥2 +1

𝑥5)3

Let 1 +1

𝑥2 +1

𝑥5 = 𝑡

(−2

𝑥3−

5

𝑥6) 𝑑𝑥 = 𝑑𝑡

(2

𝑥3+

5

𝑥6) 𝑑𝑥 = −𝑑𝑡

− ∫𝑑𝑡

𝑡3= − (

1

(−2)𝑡2) =

1

2 . 𝑡2

=1

2

1

(1 +1

𝑥2 +1

𝑥5)2 + 𝐶 =

1

2

𝑥10

(𝑥5 + 𝑥3 + 1)2+ 𝐶

= 𝑥10

2(𝑥5 + 𝑥3 + 1)2+ 𝐶

41. If the line, 𝑥−3

2=

𝑦+2

−1=

𝑧+4

3 lies in the plane, 𝑙𝑥 + 𝑚𝑦 − 𝑧 = 9, then 𝑙2 + 𝑚2 is equal to :

(1) 18

(2) 5

(3) 2

(4) 26

Solution: (3)

Solution: (1)

𝑥−3

2=

𝑦+2

−1=

𝑧+4

3 lies in 𝑙𝑥 + 𝑚𝑦 − 𝑧 = 9

The point (3, -2, -4) lies as the plane. So it should satisfy the equation of the plane.

3𝑙 − 2𝑚 + 4 = 9

3𝑙 − 2𝑚 = 5 …….(i)

The direction ratio 2, -1, 3 should be perpendicular to the line

2(𝑙) − 1. (𝑚) − 3 = 0

2𝑙 − 𝑚 = 3 …….(ii)

𝑙 = 1 and 𝑚 = −1

28

∴ 𝑙2 + 𝑚2 = 1 + 1 = 2

Hence Option [3] is correct

42. If 0 ≤ 𝑥 < 2𝜋, then the number of real values of 𝑥, which satisfy the equation cos 𝑥 + cos 2𝑥 + cos 3𝑥 +

cos 4𝑥 = 0, is :

(1) 5

(2) 7

(3) 9

(4) 3

Solution: (2)

2 cos 2𝑥 cos 𝑥 + 2 cos 3𝑥 cos 𝑥 = 0

⇒ 2 cos 𝑥 (cos 2𝑥 + cos 3𝑥) = 0

2 cos 𝑥 2 cos5𝑥

2cos

𝑥

2= 0

𝑥 =𝜋

2,3𝜋

2, 𝜋,

𝜋

5,3𝜋

5,7𝜋

5,9𝜋

5

7 Solutions.

43. The area (in sq. units) of the region {(𝑥, 𝑦): 𝑦2 ≥ 2𝑥 and 𝑥2 + 𝑦2 ≤ 4𝑥, 𝑥 ≥ 0, 𝑦 ≥ 0} is:

(1) 𝜋 −8

3

(2) 𝜋 −4√2

3

(3) 𝜋

2−

2√2

3

(4) 𝜋 −4

3

Solution: (1)

𝑦2 = 2𝑥 (Area out side the parabola)

𝑥2 + 𝑦2 ≤ 4𝑥 (Area inside the circle)

29

First finding point of intersection of the curves

𝑥2 + 𝑦2 = 4𝑥 and 𝑦2 = 2𝑥

𝑥2 + 2𝑥 = 4𝑥

𝑥2 = 2𝑥

𝑥 = 0, 𝑥 = 2

If 𝑥 = 0, then 𝑦 = 0 𝑎𝑛𝑑 𝑖𝑓 𝑥 = 2, then 𝑦 = ±2.

Co – ordinates of A(0, 0) and 𝐵(2, 2)

As 𝑥 ≥ 0 𝑦 ≥ 0 only area above 𝑥 − 𝑎𝑥𝑖𝑠 would be considered

= [∫ √4𝑥 − 𝑥2

2

0

𝑑𝑥 − √2 ∫ √𝑥 𝑑𝑥

2

0

]

= [∫ √4 − (𝑥 − 2)2

2

0

𝑑𝑥 − √2 ∫ √𝑥 𝑑𝑥

2

0

]

= [(2 − 𝑥

2) √4𝑥 − 𝑥2 +

4

2 . sin−1 (

𝑥 − 2

2)]

0

2

− √2 (𝑥

32

32

)

0

2

= [0 − 2 sin−1(−1)] − √2 .2

32√2]

[𝜋 −8

3]

44. Let �⃗�, �⃗⃗� and 𝑐 be three unit vectors such that �⃗� × (�⃗⃗� × 𝑐) =√3

2(�⃗⃗� + 𝑐). If �⃗⃗� is not parallel to 𝑐, then the angle

between �⃗� and �⃗⃗� is :

(1) 𝜋

2

(2) 2𝜋

3

30

(3) 5𝜋

6

(4) 3𝜋

4

Solution: (3)

�⃗� × (�⃗⃗� × 𝑐) =√3

2(�⃗⃗� + 𝑐)

(�⃗� . 𝑐) �⃗⃗� − (�⃗� . �⃗⃗�) 𝑐 =√3

2 �⃗⃗� +

√3

2𝑐

On comparing the coefficient of 𝑐

On both the sides.

�⃗� . �⃗⃗� = −√3

2

|�⃗�| . |�⃗⃗�| cos 𝜃 = −√3

2

cos 𝜃 = −√3

2

𝜃 =5𝜋

6

45. A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = 𝑥 units and

a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then:

(1) (4 − 𝜋)𝑥 = 𝜋𝑟

(2) 𝑥 = 2𝑟

(3) 2𝑥 = 𝑟

(4) 2𝑥 = (𝜋 + 4)𝑟

Solution: (2)

Given that 4𝑥 + 2𝜋𝑟 = 2

i.e., 2𝑥 + 𝜋𝑟 = 1

∴ 𝑟 =1−2𝑥

𝜋 …….(i)

31

Area 𝐴 = 𝑥2 + 𝜋𝑥2

= 𝑥2 +1

𝜋 (2𝑥 − 1)2

For min vale of area A

𝑑𝐴

𝑑𝑥= 0 given 𝑥 =

2

𝜋+4 …….(ii)

From (i) and (ii)

𝑟 =1

𝜋+4 …….(iii)

∴ 𝑥 = 2𝑟

46. The distance of the point (1, −5, 9) from the plane 𝑥 − 𝑦 + 𝑧 = 5 measured along the line 𝑥 = 𝑦 = 𝑧 is :

(1) 10√3

(2) 10

√3

(3) 20

3

(4) 3√10

Solution: (1)

Equation of line parallel to 𝑥 = 𝑦 = 𝑧 through

(1, −5, 9) is 𝑥−1

1=

𝑦+5

1=

𝑧−9

1= 𝜆

If 𝑃(𝜆 + 1, 𝜆 − 5, 𝜆 + 9) be point of intersection of line and plane.

⇒ 𝜆 + 1 − 𝜆 + 5 + 𝜆 + 9 = 5

⇒ 𝜆 = −10

⇒ Coordinates point are (−9, −15, −1)

⇒ Required distance = 10√3

47. If a curve 𝑦 = 𝑓(𝑥) passes through the point (1, -1) and satisfies the differential equation, 𝑦 (1 + 𝑥𝑦)𝑑𝑥 =

𝑥 𝑑𝑦, then 𝑓 (−1

2) is equal to :

32

(1) −4

5

(2) 2

5

(3) 4

5

(4) −2

5

Solution: (3)

Given differential equation

𝑦𝑑𝑥 + 𝑥𝑦2𝑑𝑥 = 𝑥𝑑𝑦

⇒ 𝑥𝑑𝑦−𝑦𝑑𝑥

𝑦2 = 𝑥𝑑𝑥

⇒ −𝑑 (𝑥

𝑦) = 𝑑 (

𝑥2

2)

Integrating we get

=𝑥

𝑦=

𝑥2

2+ 𝐶

∵ It is passes through (1, -1)

∴ 1 =1

2+ 𝐶 ⇒ 𝐶 =

1

2

∴ 𝑥2 + 1 +2𝑥

𝑦= 0 ⇒ 𝑦 =

−2𝑥

𝑥2+1

∴ 𝑓 (−1

2) =

4

5

48. If the number of terms in the expansion of (𝑎 −2

𝑥+

4

𝑥2)𝑛

, 𝑥 ≠ 0, is 28, then the sum of the coefficients of all

the terms in this expansion, is :

(1) 2187

(2) 243

(3) 729

(4) 64

Solution: (3 or Bonus)

(a −2

x+

4

𝑥2)n as the question is having three variables the total number of terms would be

(n+1)(n+2)

1.2 which is equal to 28

33

∴ (n + 1)(n + 2 ) = 56

Which gives n = 6, and sum of coefficients would be (1 − 2 + 4)6 = 36 = 729.

49. Consider 𝑓(𝑥) = tan−1 (√1+sin 𝑥

1−sin 𝑥) , 𝑥 ∈ (0,

𝜋

2). A normal to 𝑦 = 𝑓(𝑥) at 𝑥 =

𝜋

6 also passes through the point :

(1) (0,2𝜋

3)

(2) (𝜋

6, 0)

(3) (𝜋

4, 0)

(4) (0, 0)

Solution: (1)

𝑓(𝑥) = tan−1 (√1 + sin 𝑥

1 − sin 𝑥) , 𝑥 ∈ (0,

𝜋

2)

𝑓′(𝑥) = 1

1+(1+sin 𝑥

1−sin 𝑥) .

1

2 (

1+sin 𝑥

1−sin 𝑥)

−1

2 . (

cos 𝑥(1−sin 𝑥)+cos(1+sin 𝑥)

(1−sin 𝑥)2 ) at 𝑥 =𝜋

2

𝑓′(𝑥) =

122

.1

2 (

1

√3) (

√3

14

) =1

2

Slope of tangent =1

2

So slope of normal = −2

Also at 𝑥 =𝜋

6 𝑦 = tan−1 √3 =

𝜋

3.

So equation of the tangent would be (𝑦 −𝜋

3) = −2 (𝑥 −

𝜋

6)

It passes through(0,2𝜋

3)

50. For 𝑥 ∈ 𝑅, 𝑓(𝑥) = |log 2 − sin 𝑥| and 𝑔(𝑥) = 𝑓(𝑓(𝑥)), then:

(1) 𝑔′(0) = cos(log 2)

(2) 𝑔′(0) = − cos(log 2)

(3) 𝑔 is differentiable at 𝑥 = 0 and 𝑔′(0) = − sin(log 2)

(4) 𝑔 is not differentiable at 𝑥 = 0

34

Solution: (1)

In the neighborhood of 𝑥 = 0, 𝑓(𝑥) = log 2 − sin 𝑥

∴ 𝑔(𝑥) = 𝑓(𝑓(𝑥) = log 2 − sin(𝑓(𝑥))

= log 2 − 𝑠𝑜𝑛 (log 2 − sin 𝑥)

It is differentiable at 𝑥 = 0, so

∴ 𝑔′(𝑥) = − cos(log 2 − sin 𝑥) (– cos 𝑥)

∴ 𝑔′(0) = cos(log 2)

51. Let two fair six – faced dice A and B be thrown simultaneously. If 𝐸1 is the event that die A shows up four, 𝐸2

is the event that die B shows up two and 𝐸3 is the event that the sum of numbers on both dice is odd, then which

of the following statement is NOT true ?

(1) 𝐸2 and 𝐸3 are independent

(2) 𝐸1 and 𝐸3 are independent

(3) 𝐸1, 𝐸2 and 𝐸3 are independent

(4) 𝐸1 and 𝐸2 are independent

Solution: (3)

𝐸1 → 𝐴 show up 4

𝐸2 → 𝐵 shows up 2

𝐸3 → Sum is odd (i.e., even + odd or odd + even)

𝑃(𝐸1) =6

6.6=

1

6

𝑃(𝐸2) =6

6.6=

1

6

𝑃(𝐸3) =3 × 3 × 2

6.6=

1

2

𝑃(𝐸1 ∩ 𝐸2) =1

6.6= 𝑃(𝐸1) . 𝑃(𝐸2)

⇒ 𝐸1 𝑎𝑛𝑑 𝐸2 are independent

𝑃(𝐸1 ∩ 𝐸3) =1.3

6.6= 𝑃(𝐸1) . 𝑃(𝐸3)

⇒ 𝐸1 𝑎𝑛𝑑 𝐸2 are independent

𝑃(𝐸2 ∩ 𝐸3) =1.3

6.6=

1

12= 𝑃(𝐸2) . 𝑃(𝐸3)

35

⇒ 𝐸2𝑎𝑛𝑑 𝐸3 are independent

𝑃(𝐸1 ∩ 𝐸2 ∩ 𝐸3) = 0 i.e., impossible event.

52. If𝐴 = [5𝑎 −𝑏3 2

] and A adj 𝐴 = 𝐴 𝐴𝑇, then 5𝑎 + 𝑏 is equal to :

(1) 5

(2) 4

(3) 13

(4) −1

Solution: (1)

𝐴 = [5𝑎 −𝑏3 2

] and 𝐴𝑇 = [5𝑎 3−𝑏 2

]

𝐴𝐴𝑇 = [25𝑎2 + 𝑏2 15𝑎 − 2𝑏15𝑎 − 2𝑏 13

]

Now, A adj 𝐴 = |𝐴|𝐼2 = [10𝑎 + 3𝑏 0

0 10𝑎 + 3𝑏]

Given 𝐴𝐴𝑇 = 𝐴. 𝑎𝑑𝑗 𝐴

15𝑎 − 2𝑏 = 0 ……(i)

10𝑎 + 3𝑏 = 13 …….(ii)

Solving we get

5𝑎 = 2 and 𝑏 = 3

∴ 5𝑎 + 𝑏 = 5

53. The Boolean Expression (𝑝 ∧ ∼ 𝑞) ∨ 𝑞 ∨ (∼ 𝑝 ∧ 𝑞) is equivalent to :

(1) 𝑝 ∧ 𝑞

(2) 𝑝 ∨ 𝑞

(3) 𝑝 ∨∼ 𝑞

(4) ∼ 𝑝 ∧ 𝑞

Solution: (2)

(𝑝 ∧ ∼ 𝑞) ∨ 𝑞 ∨ (∼ 𝑝 ∧ 𝑞)

Set equivalent

= (𝐴 ∩ �̅�) ∪ (�̅� ∩ 𝐵) ∪ 𝐵

36

= ((𝐴 ∪ 𝐵) − (𝐴 ∩ 𝐵) ∪ 𝐵

= 𝐴 ∪ 𝐵

Hence answer is 𝑝 ∨ 𝑞.

54. The sum of all real values of 𝑥 satisfying the equation (𝑥2 − 5𝑥 + 5)𝑥2+4𝑥−60 = 1 is:

(1) - 4

(2) 6

(3) 5

(4) 3

Solution: (4)

(𝑥2 − 5𝑥 + 5)𝑥2+4𝑥−60 = 1

(𝑥2 − 5𝑥 + 5)(𝑥+10) (𝑥−6) = 1

𝑥 = −10 and 𝑥 = 6 will make L.H.S = 1.

Also at 𝑥 = 1; (1)11.(−5) = 1

And at 𝑥 = 4; (1)14.(−2) = 1

We should also considered the case when 𝑥2 − 5𝑥 + 5 = −1, and it has even power

𝑥2 − 5𝑥 + 6 = 0

(𝑥 − 2)(𝑥 − 3) = 0

So 𝑥 = 2 will give = (−1)𝑒𝑣𝑒𝑛

At 𝑥 = 2

(−1)12(−4) = 1

So sum would be

−10 + 6 + 1 + 4 + 2 = 3

Hence answer is 3

55. The centres of those circles which touch the circle, 𝑥2 + 𝑦2 − 8𝑥 − 8𝑦 − 4 = 0, externally and also touch the x

– axis, lie on :

(1) An ellipse which is not a circle

(2) A hyperbola

37

(3) A parabola

(4) A circle

Solution: (3)

𝑥2 + 𝑦2 − 8𝑥 − 8𝑦 − 4 = 0

has centre (4, 4) and radius 6.

Let (ℎ, 𝑘) be the centre of the circle which is touching the circle externally

Then

√(ℎ − 4)2 + (𝑘 − 4)2 = 6 + 𝑘

ℎ2 − 8ℎ + 16 + 𝑘2 − 8𝑘 + 16 = 36 + 12𝑘 + 𝑘2

ℎ2 − 8ℎ − 20𝑘 − 4 = 0,

Replacing ℎ by 𝑥 and 𝑘 by 𝑦

𝑥2 − 8𝑥 − 20𝑦 − 4 = 0

Equation of parabola.

56. If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and

arranged as in a dictionary; then the position of the word SMALL is :

(1) 59𝑡ℎ

(2) 52𝑛𝑑

(3) 58𝑡ℎ

(4) 46𝑡ℎ

Solution: (3)

SMALL

Total number of words formed would be ∟5

∟2= 60

When arranged as per dictionary. The words starting from A

𝐴 − − − −∟4

∟2= 12

The words standing from 𝐿

𝐿 − − − − = ∟4 = 24

The words starting form M

38

𝑀 − − − − =∟4

∟2= 12

The words starting from

𝑆 𝐴 − − − =∟3

∟2= 3

𝑆 𝐿 − − − ∟3 = 6

S M A L L = 1

Rank would be 12 + 24 + 12 + 3 + 6 + 1 = 58

57. lim𝑛→∞

(𝑛+1) (𝑛+2)….3𝑛

𝑛2𝑛 )

1

𝑛 is equal to :

(1) 27

𝑒2

(2) 9

𝑒2

(3) 3 log 3 − 2

(4) 18

𝑒4

Solution: (1)

lim𝑛→∞

((𝑛 + 1) (𝑛 + 2) … .3𝑛

𝑛2𝑛 )

1𝑛

Let 𝑦 = ((𝑛+1) (𝑛+2)…..2𝑛+1)

𝑛2𝑛 )

1

𝑛

𝑦 = ((𝑛 + 1)

𝑛 .

(𝑛 + 2)

𝑛… …

(2𝑛 + 𝑛)

𝑛)

1𝑛

log 𝑦 =1

𝑛[log (1 +

1

𝑛) + log (1 +

2

𝑛) + ⋯ . . log(1 + 2)]

As 𝑛 → ∞

log 𝑦 = ∫ log(1 + 𝑥) 𝑑𝑥

2

0

Integrating by parts

log 𝑦 = ∫ 1. log(1 + 𝑥) 𝑑𝑥

2

0

39

= (𝑥 . log(1 + 𝑥))02 − ∫

1

1 + 𝑥

2

0

𝑥

= (𝑥 log(1 + 𝑥))02 − ∫ (

1 + 𝑥

1 + 𝑥)

2

0

+ ∫1

1 + 𝑥

2

0

= (𝑥 log(1 + 𝑥))02 − (𝑥)0

2 + (log(1 + 𝑥))02

= (2 log 3 − 0) − (2 − 0) + (log 3 − log 1)

= 3 log 3 − 2

Since log 𝑦 = 3 log 3 − 2

= 𝑦 =𝑒log 27

𝑒2=

27

𝑒2

=27

𝑒2

58. If the sum of the first ten terms of the series (13

5)

2+ (2

2

5)

2+ (3

1

5)

2+ 42 + (4

4

5)

2+ ⋯ ., is

16

5𝑚, then 𝑚 is

equal to :

(1) 101

(2) 100

(3) 99

(4) 102

Solution: (1)

(8

5)

2+ (

12

5)

2+ (

16

5)

2……..10 terms

𝑇𝑛 = (4𝑛 + 4

5)

2

𝑇𝑛 = 16 (𝑛2 + 2𝑛 + 1

25)

𝑇𝑛 =16

25(42 + 2𝑛 + 1)

𝑇𝑛 = 𝑆𝑛 = (16

25) (

𝑛(𝑛 + 1) (2𝑛 + 1)

6+

2(𝑛) (𝑛 + 1) + 4

2)

40

Put 𝑛 = 10

16

25(

10 . 11. 21

6) +

2 . 10 . 11

2+ 10

=16

25 (385 + 110 + 10) =

16

25 . 505

=16

5 101.

Hence m = 101

59. If one of the diameters of the circle, given by the equation, 𝑥2 + 𝑦2 − 4𝑥 + 6𝑦 − 12 = 0, is a chord of a circle

S, whose centre is at (-3, 2), then the radius of S is:

(1) 5√3

(2) 5

(3) 10

(4) 5√2

Solution: (1)

The centre of the given circle 𝑥2 + 𝑦2 − 4𝑥 + 6𝑦 − 12 = 0 is (2, -3) and the radius is 5.

The distance between the centres 5√2 and radius is 5. The triangle OPQ is a right angled triangle

𝑂𝑄 = √(5√2)2

+ 52 = √(5√3)2

= 5√3

Hence answer is 5√3

60. A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the

path, he observes that the angle of elevation of the top of the pillar is 30𝑜. After walking for 10 minutes from A in

the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60𝑜. Then the

time taken (in minutes) by him, from B to reach the pillar, is :

(1) 10

41

(2) 20

(3) 5

(4) 6

Solution: (3)

Δ𝑄𝑃𝐴 ∶ℎ

𝑥+𝑦= tan 30𝑜 ⇒ √3ℎ = 𝑥 + 𝑦 ……(i)

Δ𝑄𝑃𝐵 ∶ℎ

𝑦= tan 60𝑜 ⇒ ℎ = √3𝑦 ……(ii)

By (i) and (ii) : 3𝑦 = 𝑥 + 𝑦 ⇒ 𝑦 =𝑥

2

∵ Speed is uniform

Distance 𝑥 in 10 mins

⇒ Distance 𝑥

2 in 5 mins.

PHYSICS

61. A uniform string of length 20m is suspended from a rigid support. A short wave pulse is introduced at its

lowest end. It starts moving up the string. The time taken to reach the support is:

(take g = 10 ms−2)

(1) 2𝑠

(2) 2√2 𝑠

(3) √2 𝑠

(4) 2𝜋√2 𝑠

Solution: (2)

𝑡 = 2√𝑙

𝑔= 2√2 second.

62. A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1m 1000 times. Assume

that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up

42

considering the work done only when the weight is lifted up? Fat supplies 3.8 × 107 J of energy per kg which

is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms−2:

(1) 6.45 × 10−3 kg

(2) 9.89 × 10−3 kg

(3) 12.89 × 10−3 kg

(4) 2.45 × 10−3 kg

Solution: (3)

m = 10kg, h = 1m, 1000 times

PE = 98 J × 1000 = 98000 J = 98 kJ

= 9.8 × 104 J

Fat burn = 3.8 × 107 J × 0.2

= 7.6 × 106 J per kg

m =9.8 × 104

7.6 × 106= 1.289 × 10−2

= 12.89 × 10−3 kg

63. A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient

of friction, between the particle and the rough track equals μ. The particle is released, from rest, from the point P

and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal

to each other, and no energy is lost when particle changes direction from PQ to QR.

The values of the coefficient of friction μ and the distance x = (QR), are respectively close to:

(1) 0.2 and 3.5 m

(2) 0.29 and 3.5 m

(3) 0.29 and 6.5 m

(4) 0.2 and 6.5 m

Solution: (2)

43

tan 30o =h

l

l = h√3 = 2√3 m

Wf = −μmgl or Wf = −μmgx

μmgl = μmgx; x = l

x = 2√3 m; Wall = ∆K

mgh − μmgl − μmgx = 0

h − μl − μx = 0

2 = μ(l + x) ⇒ μ =2

l + x=

2

4√3=

1

2√3

64. Two identical wires A and B, each of length ‘𝑙’, carry the same current 𝐼. Wire A is bent into a circle of radius

R and wire B is bent to form a square of side ‘a’. If BA and BB are the values of magnetic field at the centres of the

circle and square respectively, then the ratio BA

BB is :

(1) π2

16√2

(2) π2

16

(3) π2

8√2

(4) π2

8

Solution: (3)

2πR = 4a

a

R=

2π

4 a

R=

π

2

44

BA =μ0i

2R

BB =μ0i

πa(2√2)

BA

BB=

μ0i

2R×

πa

2√2 μ0i

=πa

4√2R=

π

4√2(

π

2) =

π2

8√2

65. A galvanometer having a coil resistance of 100 Ω gives a full scale deflection, when a current of 1 mA is

passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full

scale deflection for a current of 10 A, is:

(1) 2 Ω

(2) 0.1 Ω

(3) 3 Ω

(4) 0.01 Ω

Solution: (4)

We know that

RS =IG

1 − IGRG

=1 × 10−3

10× 100

= 10−2

= 0.01 Ω

66. An observer looks at a distant tree of height 10m with a telescope of magnifying power of 20. To the

observer the tree appears:

(1) 10 times nearer.

(2) 20 times taller.

(3) 20 times nearer.

(4) 10 times taller.

45

Solution: (2)

θ =10

x

θ1 =10

x(20)

Now 20 times Taller.

67. The temperature dependence of resistance of Cu and undoped Si in the temperature range 300-400 K, is

best described by:

(1) Linear increase for Cu, exponential increase for Si.

(2) Linear increase for Cu, exponential decrease for Si.

(3) Linear decrease for Cu, linear decrease for Si

(4) Linear increase for Cu, linear increase for Si.

Solution: (2)

Resistance variation with temperature: Cu-metal, undoped Silicon-Semi Conductor resistance of metal

increases with increase in temperature linearly resistance of semi Conductor decreases exponentially with

increase in temperature.

68. Choose the correct statement:

(1) In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the

amplitude of the audio signal.

(2) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the

amplitude of the audio signal.

(3) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the

frequency of the audio signal.

(4) In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the

amplitude of the audio signal.

Solution: (1)

As per properties of A.M. in amplitude modulation the amplitude of high frequency carrier wave is made to vary in

proportion to the amplitude of the audio signal.

46

69. Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the

samples have equal number of nuclei. After 80 minutes, the ratio of decayed numbers of A and B nuclei will be:

(1) 4 : 1

(2) 1 : 4

(3) 5 : 4

(d) 1 : 16

Solution: (3)

t = 80 min = 4 TA = 2TB

no. of nuclei of A decayed = N0 −N0

24 =15N0

16

no. of nuclei of B decayed = N0 −N0

22 =3N0

4

required ratio =5

4

70. ‘n’ moles of an ideal gas undergoes a process A → B as shown in the figure. The maximum temperature of

the gas during the process will be:

(1) 3 P0V0

2nR

(2) 9 P0V0

2nR

(3) 9 P0V0

nR

(4) 9 P0V0

4nR

Solution: (4)

Tmax at mid point

T =pv

nR=

(32 P0) (

3V02 )

nR

47

=9

4(

P0V0

nR)

71. An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz

AC supply, the series inductor needed for it to work is close to:

(1) 0.08 H

(2) 0.044 H

(3) 0.065 H

(4) 80 H

Solution: (3)

VL2 + 6400 = 220 × 220

IR = 80

VL = √48400 − 6400

I =80

8= 10 = √42000 = 210

I XL = 210

XL = 2πfL = 210

L =210

10 × 100 π= 0.065 H

72. A pipe open at both ends has a fundamental frequency 𝑓 in air. The pipe is dipped vertically in water so that

half of it is in water. The fundamental frequency of the air column is now:

(1) 3𝑓

4

(2) 2𝑓

(3) 𝑓

(4) 𝑓

2

48

Solution: (3)

f =v

2l

f ′ =v

4l′=

v

4 (l2

)= f

73. The box of a pin hole camera, of length L has a hole of radius a. It is assumed that when the hole is

illuminated by a parallel beam of light of wavelength λ the spread of the spot (obtained on the opposite wall of the

camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its

minimum size (say bmin) when:

(1) a = √λL and bmin = (2λ2

L)

(2) a = √λL and bmin = √4λL

(3) a =λ2

L and bmin = √4λL

(4) a =λ2

L and bmin = (

2λ2

L)

Solution: (2)

The diffraction angle λa cause a spreading of Lλ

a in the size of the spot. These become large when a

(Radius) is small.

So adding of two kind of spreading (for simplicity) we get spot size is

a +La

a.

Hence to find out minimum value of this

We can write it as √(a −Lλ

a)

2+ 4Lλ

49

∵ The minimum value is when a =La

a i.e. the geometric and diffraction broadening are equal √4Lλ

∴ When a = √λL and bmin = √4λL

74. A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point

charge Q (having a charge equal to the sum of the charges on the 4 μF and 9 μF capacitors), at a point distant 30

m from it, would equal:

(1) 360 N/C

(2) 420 N/C

(3) 480 N/C

(4) 240 N/C

Solution: (2)

Potential at 4μF = 6 volt

∴ charge q1 = 24μC

Potential at 9μF = 2 volt

∴ charge q2 = 18 μC

Total q = 42 μC

E =kq

r2 =9×109×42×10−6

900= 420 N/C

75. Arrange the following electromagnetic radiations per quantum in the order of increasing energy:

A: Blue light

B: Yellow light

50

C: X-ray

D: Radiowave

(1) A, B, D, C

(2) C, A, B, D

(3) B, A, D, C

(4) D, B, A, C

Solution: (4)

76. Hysteresis loops for two magnetic materials A and B are given below:

These materials are used to make magnets for electric generators, transformer core and electromagnet core.

Then it is proper to use:

(1) A for electromagnets and B for electric generators.

(2) A for transformers and B for electric generators.

(3) B for electromagnets and transformers.

(4) A for electric generators and transformers.

Solution: (3)

Graph A is hard ferromagnetic material substance.

51

The graph of B is graph of soft ferromagnetic material which is we use to consist of electromagnets and

transformers.

77. A pendulum clock loses 12s a day if the temperature is 40o C and gains 4s a day if the temperature is 20o

C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion (α) of

the metal of the pendulum shaft are respectively:

(1) 60oC; α = 1.85 × 10−4/ oC

(2) 30oC; α = 1.85 × 10−3/ oC

(3) 55oC; α = 1.85 × 10−2/ oC

(4) 25oC; α = 1.85 × 10−5/ oC

Solution: (4)

∆T ∝ ∆θ

12

4=

40 − θ

θ − 20

3θ − 60 = 40 − θ

4θ = 100

θ = 25oC

∆T =1

2α ∆θ × T

4 =1

2α 5 × 86400;

8×105

5×86400= α;

8000

4320= α

α = 1.05 × 10−5/ oC

78. The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see figure), has volume

charge density ρ =A

r, where A is a constant and r is the distance from the centre. At the centre of the spheres

52

is a point charge Q. The value of A such that the electric field in the region between the spheres will be

constant, is:

(1) Q

2π(b2−a2)

(2) 2Q

π(a2−b2)

(3) 2Q

πa2

(4) Q

2πa2

Solution: (4)

E =K [Q + ∫ 4πx2dx

Ax

r

a]

r2

E = K [Q

r2+ 4μA {

r2

2r2−

a2

2r2}]

dE

dr= 0

Q

r2+

4πAa2

2r2

A =Q

2πa2

79. In an experiment for determination of refractive index of glass of a prism by i = δ, plot, it was found that a

ray incident at angle 35o, suffers a deviation of 40o and that it emerges at angle 79o. In that case which of the

following is closest to the maximum possible value of the refractive index?

53

(1) 1.6

(2) 1.7

(3) 1.8

(4) 1.5

Solution: (4)

i = 35o, δ = 40o, e = 79o

δ = i + e − A

40o = 35o + 79o − A

A = 74o

And r1 + r2 = A = 74o

Solving these, we get μ = 1.5

Since δmin < 40o

μ <sin (

74 + 402 )

sin 37

μmax = 1.44

80. A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is

90s, 91s, 95s and 92s. If the minimum division in the measuring clock is 1s, then the reported mean time

should be:

(1) 92 ± 5.0𝑠

(2) 92 ± 1.8𝑠

(3) 92 ± 3𝑠

(4) 92 ± 2𝑠

Solution: (4)

T Ts Ti − T (Ti − T)2

t1 90 -2 4

t2 91 -1 1

t3 95 3 9

t4 92 0 0

Ti 92 ΣTi − 𝑇

N= 0

Σ(Ti − T)2

N= 3.5

Tr = 𝑇 ± √Σ(Ti − T )2

N

Tr = 92 ± √3.5

Tr = 92 ± 1.8

54

Tr = 92 ± 2

Because least count of clock is 1s.

81. Identify the semiconductor devices whose characteristics are given below, in the order (a), (b), (c), (d):

(1) Zener diode, Simple diode, Light dependent resistance, Solar cell

(2) Solar cell, Light dependent resistance, Zener diode, Simple diode

(3) Zener diode, Solar cell, Simple diode Light dependent resistance

(4) Simple diode, Zener diode, Solar cell, Light dependent resistance

Solution: (4)

Its V-I characteristics of simple diode.

55

Its V-I characteristic of Zener diode.

V-I characteristics of Solar cell

V-I characteristics of light dependence resistance.

82. Radiation of wavelength λ, is incident on a photocell. The fastest emitted electron has speed 𝜐. If the

wavelength is changed to 3λ

4, the speed of the fastest emitted electron will be:

(1) < 𝜐 (4

3)

1

2

(2) = 𝜐 (4

3)

1

2

(3) = 𝜐 (3

4)

1

2

(4) > 𝜐 (4

3)

1

2

Solution: (4)

1

2𝑚𝑣2 =

ℎ𝑐

𝜆− 𝜙 …(i)

1

2𝑚𝑣′2 =

4

3

ℎ𝑐

𝜆− 𝜙 …(ii)

56

From eqn. (i) ℎ𝑐

𝜆=

1

2𝑚𝑣2 + 𝜙

On putting this equ. (ii)

1

2𝑚𝑣′2 =

4

3(

1

2𝑚𝑣2 + 𝜙) − 𝜙

𝑣′ > 𝑣√4

3

83. A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a

distance 2A

3 from equilibrium position. The new amplitude of the motion is:

(1) 3𝐴

(2) 𝐴√3

(3) 7𝐴

3

(4) 𝐴

3√41

Solution: (3)

v = ω√A2 − X2 ;

v = ω√A2 −4A2

9

=ω√5A

3

New SHM will be,

3v − ω√AN2 − Xn

2;

3ω√5A

3= ω√AN

2 −4A2

9

5A2 = AN2 −

4A2

9

AN2 =

49A2

9

AN =7A

3

57

84. A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed υ in the x-y plane

as shown in the figure:

Which of the following statements is false for the angular momentum L⃗⃗ about the origin?

(1) L⃗⃗ = mυ [R

√2− a] k̂ when the particle is moving from C to D.

(2) L⃗⃗ = mυ [R

√2+ a] k̂ when the particle is moving from B to C.

(3) L⃗⃗ =mυ

√2Rk̂ when the particle is moving from D to A.

(4) L⃗⃗ = −mυ

√2Rk̂ when the particle is moving from A to B.

Solution: (1, 3)

L⃗⃗ = r⃗ × P⃗⃗⃗ or L⃗⃗ = rp sin θ n̂

Or L⃗⃗ = r⊥(P)n̂

For D to A, L⃗⃗ =R

√2mV(−k̂)

For A to B, L⃗⃗ =R

√2mV(−k̂)

For C to D, L⃗⃗ = (R

√2+ a) mV(k̂)

For B to C, L⃗⃗ = (R

√2+ a) mV(k̂)

85. An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains

constant. If during this process the relation of pressure P and volume V is given by PVn = constant, then n is

given by (Here CP and CV are molar specific heat at constant pressure and constant volume, respectively) :

(1) n =C−CP

C−CV

58

(2) n =CP−C

C−CV

(3) n =C−CV

C−CP

(4) n =CP

CV

Solution: (1)

PVn = k

C = Cv +R

1−n; C − Cv =

R

1−n

1 − n =R

C−Cv; n = 1 −

R

C−Cv

n =C−Cv−R

C−Cv ; n =

C−Cv−(Cp−Cv)

C−Cv

n =C−Cv−Cp+Cv

C−Cv; n =

C−Cp

C−Cv

86. A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness

of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw

gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main

scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division

coincides with the main scale line?

(1) 0.80 mm

(2) 0.70 mm

(3) 0.50 mm

(4) 0.75 mm

Solution: (1)

LC =0.5

50= 0.01 mm

Zero error = 0.50 − 0.45 = −0.05

Thickness = (0.5 + 25 × 0.01) + 0.05

= 0.5 + 0.25 + 0.05

= 0.8 mm

87. A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD which are

placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining

AB and CD (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the

direction shown. As it moves, the roller will tend to:

59

(1) to right.

(2) go straight.

(3) turn left and right alternately.

(4) turn left.

Solution: (4)

Say the distance of central line from instantaneous axis of rotation is r.

Then r from the point on left becomes lesser than that for right.

So Vleft point = ωr′ < ωr = vright point

So the roller will turn to left.

88. If a, b, c are inputs to a gate and x is its output, then, as per the following time graph, the gate is:

(1) AND

(2) OR

(3) NAND

(4) NOT

Solution: (2)

a b c d X

0 0 0 0 0

60

0 0 0 1 1

When any input is one output is one hence the gate is ‘OR’ gate.

89. For a common emitter configuration, if α and β have their usual meanings, the incorrect relationship between

α and β is:

(1) α =β

1−β

(2) α =β

1+β

(3) α =β2

1+β2

(4) 1

α=

1

β+ 1

Solution: (1, 3)

IE = IC + IB;

IE

IC= 1 +

IB

IC

1

∝= 1 +

1

β;

1

∝=

β+1

β

∝=β

1 + β

90. A satellite is revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth R; h << R).

The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s

gravitational field, is close to: (Neglect the effect of atmosphere.)

(1) √gR

(2) √𝑔𝑅

2

(3) √𝑔𝑅 (√2 − 1)

(4) √2𝑔𝑅

Solution: (3)

Since ℎ << 𝑅

V0 = √2gR

& Ve = √gR

∴ min velocity required

V0 − Ve = √2gR − √gR

= (√2 − 1)√gR